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I have a problem. I have tried to solve it but I get $125/6$ instead of $9/2$ (textbook result).

Find the area of the parts of plane given by the solutions of the following system: $$ \left\{\begin{matrix} y\geq x^2-4x\\ x+y-4\leq 0 \end{matrix}\right.$$

graph I realized that I needed the area of the parabolic segment. I have found the intersections A and B of the line with parabola, then I got the length of this segment, $5\sqrt2$.

I have calculated the line parallel to $x+y-4=0$ and tangent to the hyperbola: $y=-x-9/4$.

Then I have calculated the distance of the point B from the line and I got $\frac{25}{4\sqrt{2}}$.

So I could calculate the area of the rectangle and then multiply it by $2/3$ (I want the area of the parabolic segment). I got $125/6$ but my textbook says the result is $9/2$. Where's the mistake?

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According to the statement of the question, shouldn't you just calculate $\int_A^B(4-x)-(x^2-4x)dx$? –  Easy Apr 4 '13 at 8:40
    
@Easy but I'm not supposed to know how to manage integrals and derivatives, so I have to use analytic geometry! –  Surfer on the fall Apr 4 '13 at 9:37
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